According to this question:

What determines color -- wavelength or frequency?

It is the energy, and thus the frequency of a photon $E=hf$, that determines where it lies in the electromagnetic spectrum (for example its color). Wavelength is determined by the index of refraction of the medium it is travelling through via $λ=v/f$.

Does this mean that a photon would appear red in color regardless of how short or long its wavelength is, so long as its energy corresponds with that of red light??? Does this mean an x-ray or gamma-ray or radio wave is still what it was initially despite of a substantially different wavelength?

For example if you pass a red light through something with an extremely high $n$ such that $v=1 \times 10^{-6}\:\mathrm{m/s}$....

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    $\begingroup$ A photon appears red due to the interaction with the cells in your retina. Those have a fixed $n$, so the wavelength received in your eyes will not vary. But even if it did, opsin molecules are small compared to the wavelength of visible light, so if you increased the wavelength, not much would change in the interaction with light, as the frequency of the light would still drive the same transitions. $\endgroup$ – Sebastian Riese Oct 17 '15 at 16:44
  • $\begingroup$ I had seen that question before posting, hence referencing it in the question... What I wanted to know that I wasn't getting from the answers there is whether an x-ray is still an x-ray if it has an extremely long wavelength or whether red light is still red light if it has an extremely short wavelength so long as its energy is the same. That is, will it still behave the same way?? Can an x-ray with a really long wavelength still produce those images where it shows the bones in your body...or is there a wavelength where it stops being capable of this while still being technically an x-ray. $\endgroup$ – Jet Blue Oct 17 '15 at 17:59
  • $\begingroup$ Your equation $v=\lambda/f$ is wrong. $\endgroup$ – Bill N Oct 17 '15 at 18:11
  • $\begingroup$ Good call, will edit that. Question still stands tho $\endgroup$ – Jet Blue Oct 17 '15 at 18:14
  • $\begingroup$ @JetBlue The answers of the other question answer your question as well (and they are qualitatively the same question, you only add the number). $\endgroup$ – Sebastian Riese Oct 17 '15 at 19:16

Firstly, $E=hf$ is always true for electromagnetic radiation like light. In vacuum the speed of light is always $c$, so that in vacuum:

$c=\lambda f$.

In any transparent medium, other than vacuum, light slows down to $v<c$ and we define the refractive index $n$ of that medium as:


With $E=hf$ and (in a medium other than vacuum) $v=\lambda f$, then:

$\large{\lambda=\frac{vh}{E}}$ and since as $v < c$, wavelength is reduced in a medium other than vacuum.

Does this mean that a photon would appear red in color regardless of how short or long its wavelength is, so long as its energy corresponds with that of red light???

Yes, the perceived colour is only affected by the energy of the light, not by the (colourless and transparent) medium through which it travels. That is why light refracted by a prism changes direction (at the interface air/prism) but maintains its colour.

Slowing down a light pulse to the speed of a bicycle, using a Bose-Einstein Condensate (video).

Photons can also be identified by their wavelengths, using a diffraction grating.

  • $\begingroup$ Does this apply to any member of the EM spectrum? Will an x-ray with really really long wavelengths still be able to produce the x-ray images where you can see bones... or does it lose this ability and others as the wavelength changes... $\endgroup$ – Jet Blue Oct 17 '15 at 18:03
  • $\begingroup$ What is your definition of an x-ray? How do you propose to get an x-ray with a very long wavelength? $\endgroup$ – Bill N Oct 17 '15 at 18:06
  • $\begingroup$ E=hf .... so with energy that corresponds to x-ray $\endgroup$ – Jet Blue Oct 17 '15 at 18:12
  • $\begingroup$ You can't make the wavelength longer. Then it would have $v>c$. $\endgroup$ – Bill N Oct 17 '15 at 18:14
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    $\begingroup$ @JetBlue: for use in radiography it would not matter much because we use X-rays because of their penetrative power, not their wavelength. But in electron microscopy the smaller wavelength would actually be an advantage because the wavelength defines the resolution of the microscope. $\endgroup$ – Gert Oct 17 '15 at 18:36

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